如果我这样做:
result = reduce(operator.and_, [False] * 1000)
它会在第一个结果后停止吗? (因为 False & anything == False
)
类似地:
result = reduce(operator.or_, [True] * 1000)
最佳答案
事实并非如此。在这种情况下,您的替代方案是 any和 all .
result = reduce(operator.and_, [False] * 1000)
result = reduce(operator.or_, [True] * 1000)
可以替换为
result = all([False] * 1000)
result = any([True] * 1000)
短路。
计时结果显示差异:
In [1]: import operator
In [2]: timeit result = reduce(operator.and_, [False] * 1000)
10000 loops, best of 3: 113 us per loop
In [3]: timeit result = all([False] * 1000)
100000 loops, best of 3: 5.59 us per loop
In [4]: timeit result = reduce(operator.or_, [True] * 1000)
10000 loops, best of 3: 113 us per loop
In [5]: timeit result = any([True] * 1000)
100000 loops, best of 3: 5.49 us per loop
关于python - Python的reduce()会短路吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3570624/